Question

Solve for the values of x in the equation: 2^(x) = 4x.

Answer 1

There are two of them. 

I don't know a mechanical way to 'solve' for them.

One can be found by trial and error:

x=0 . . . . . 2^0 = 1 . . . . . 4(0) = 0 . . . . . no, that doesn't work
x=1 . . . . . 2^1 = 2 . . . . . 4(1) = 4 . . . . . no, that doesn't work
x=2 . . . . . 2^2 = 4 . . . . . 4(2) = 8 . . . . . no, that doesn't work
x=3 . . . . . 2^3 = 8 . . . . . 4(3) = 12 . . . . no, that doesn't work
x=4 . . . . . 2^4 = 16 . . . . 4(4) = 16 . . . . Yes !  That works !       yay !

For the other one, I constructed tables of values for 2^x and (4x)
in a spread sheet, then graphed them, and looked for the point
where the graphs of the two expressions cross.

The point is near, but not exactly,         x = 0.30990693...

If there's a way to find an analytical expression for the value, it must involve
some esoteric kind of math operations that I didn't learn in high school or
engineering school, and which has thus far eluded me during my lengthy
residency in the college of hard knocks.  If anybody out there has it, I'm
waiting with all ears.

Answer 2

$2^{x} = 4x \\ln(2^{x}) = ln(4x) \\\frac{xln(2)}{ln(2)} = \frac{ln(4x)}{ln(2)} \\x = 4$